In the decay sequence,
`{:(238),(92):}Uoverset(-x_(1))rarr{:(234),(90):}"Th"overset(-x_(2))rarr{:(234),(91):}"Pa"overset(-x_(3))rarr234_(Z) overset(-x_(4))rarr{:(230),(90):}`
`x_(1),x_(2),x_(3) and x_(4)` are particles/radiation emitted by the respective isotopes. The correct option(s) is (are):

Options

A. `x_(2)" is "beta^(-)`
B. Z is an isotope of uranium
C. `x_(3)` is x-ray
D. `x_(1)` will deflect towards negatively charged plate

Explanation

To solve the problem regarding the decay sequence of Uranium-238, we need to identify the particles emitted during each decay step. Let's break down the sequence step by step.

### Step 1: Identify the first decay
The first decay is from Uranium-238 (U) to Thorium-234 (Th):
- **Uranium-238**: Atomic number = 92, Mass number = 238
- **Thorium-234**: Atomic number = 90, Mass number = 234

In this decay, we can see that the atomic number decreases by 2 and the mass number decreases by 4. This indicates the emission of an alpha particle (α), which consists of 2 protons and 2 neutrons.

**X1 = Alpha particle (α)**

### Step 2: Identify the second decay
The second decay is from Thorium-234 (Th) to Protactinium-234 (Pa):
- **Thorium-234**: Atomic number = 90, Mass number = 234
- **Protactinium-234**: Atomic number = 91, Mass number = 234

Here, the atomic number increases by 1 while the mass number remains the same. This indicates the emission of a beta particle (β), which is a high-energy, high-speed electron or positron.

**X2 = Beta particle (β)**

### Step 3: Identify the third decay
The third decay is from Protactinium-234 (Pa) to an isotope Z:
- **Protactinium-234**: Atomic number = 91, Mass number = 234
- **Isotope Z**: Atomic number = Z, Mass number = 234

In this case, the atomic number increases by 1 again, indicating another beta decay.

**X3 = Beta particle (β)**

### Step 4: Identify the fourth decay
The fourth decay is from isotope Z to Thorium-230 (Th):
- **Isotope Z**: Atomic number = Z, Mass number = 234
- **Thorium-230**: Atomic number = 90, Mass number = 230

Here, the atomic number decreases by 2 and the mass number decreases by 4, indicating the emission of another alpha particle (α).

**X4 = Alpha particle (α)**

### Summary of emissions:
- **X1** = Alpha particle (α)
- **X2** = Beta particle (β)
- **X3** = Beta particle (β)
- **X4** = Alpha particle (α)

### Conclusion
The correct options for the particles emitted are:
- X1 = α
- X2 = β
- X3 = β
- X4 = α